Question

A plastic-epoxy sheet has uniform holes of radius \(1.99 \mathrm{~cm}\). The holes are intended to allow solid ball bear- ings with an outer radius of \(2.00 \mathrm{~cm}\) to just go through. Over what temperature rise must the plastic-epoxy sheet be heated so that the ball bearings will go through the holes? The linear expansion coefficient of plastic-epoxy is about \(1.3 \cdot 10^{-4}{ }^{\circ} \mathrm{C}^{-1}\).

Short answer

Answer: The minimum temperature increase required is 38.16°C.

Step-by-step solution

Identifying given parameters

First, we need to identify the given parameters. The initial hole radius in the plastic-epoxy sheet is \(1.99\) cm, and the outer radius of the ball bearing is \(2.00\) cm. The linear expansion coefficient of the plastic-epoxy is given as \(1.3 \cdot 10^{-4}{ }^{\circ} \mathrm{C}^{-1}\). We need to find the temperature increase required for the ball bearing to pass through the hole.

Setting up the expansion equation

The expansion equation for the hole in the plastic-epoxy sheet can be written as: \(\Delta r = \alpha \cdot r_o \cdot \Delta T\) where \(\Delta r\) is the change in radius of the hole, \(\alpha\) is the linear expansion coefficient, \(r_o\) is the initial hole radius, and \(\Delta T\) is the temperature change. We want the radius of the hole, \(r_f\) to be at least equal to the radius of the ball bearing. So, the equation becomes: \(r_f \geq r_o + \alpha \cdot r_o \cdot \Delta T\)

Calculating the necessary change in radius

We need to find the required change in radius, \(\Delta r\), for the hole to accommodate the ball bearing. We can calculate this as follows: \(\Delta r = r_{\text{ball bearing}} - r_o\) \(\Delta r = 2.00 - 1.99\) \(\Delta r = 0.01 \mathrm{~cm}\)

Solving for the required temperature increase

Now we can substitute the values into the equation for the expansion of the hole and solve for the temperature increase, \(\Delta T\). Rearrange the equation to isolate \(\Delta T\): \(\Delta T \geq \frac{\Delta r}{\alpha \cdot r_o}\) Substitute the given values: \(\Delta T \geq \frac{0.01}{1.3 \cdot 10^{-4} \cdot 1.99}\) Now, calculate the temperature change: \(\Delta T \geq 38.16^{\circ} \mathrm{C}\)

Final answer

The plastic-epoxy sheet must be heated by at least \(38.16^{\circ} \mathrm{C}\) for the ball bearings with an outer radius of \(2.00 \mathrm{~cm}\) to go through the holes.

Key concepts

Linear Expansion Coefficient

Understanding the linear expansion coefficient is essential for solving problems regarding the change in dimensions of materials due to temperature variations. The linear expansion coefficient, denoted by \(\alpha\), measures how much a material's linear dimension changes per degree change in temperature. It's a fundamental property expressing the tendency of materials to increase in length when heated. Different materials expand at different rates, with this coefficient commonly expressed in units of inverse temperature \( ^{\circ}C^{-1} \) or \(K^{-1}\).

In the case of the plastic-epoxy sheet exercise, \(\alpha\) for the plastic-epoxy is given as \(1.3 \cdot 10^{-4}{ }^{\circ} C^{-1}\), a relatively high value indicating that this material expands significantly with temperature increase. Such knowledge is crucial in applications where precision in mechanical fit is of the essence, such as in the fit between the ball bearings and the holes in the sheet.

Temperature Change Calculation

To determine the necessary temperature change for materials to expand or contract to desired dimensions, we use the temperature change calculation. This calculation relies on the formula for linear expansion \(\Delta r = \alpha \cdot r_o \cdot \Delta T\), where \(\Delta r\) is the change in length, \(\alpha\) is the linear expansion coefficient, \(r_o\) is the original length, and \(\Delta T\) is the change in temperature.

When we need the material to fit a certain size, we rearrange the equation to solve for \(\Delta T\). In our context, ensuring the ball bearings will fit through the heated holes involves finding the minimum temperature increase, as shown in the steps provided. Through rearrangement and substitution of known values, we calculate \(\Delta T\). Paying close attention to dimensional accuracy and unit consistency ensures the reliability of the final result.

Physical Properties of Materials

The performance and suitability of materials for specific applications depend on their distinct physical properties, which determine how they respond to environmental factors like temperature. Along with linear expansion coefficient, other properties include density, electrical conductivity, thermal conductivity, and specific heat capacity. Each property can greatly influence how a material is used and how it should be handled in different scenarios.

A robust understanding of these physical properties is paramount. For instance, when selecting materials for high-temperature environments, one must consider not just the linear expansion but also the material's ability to retain strength at elevated temperatures. In our plastic-epoxy sheet scenario, the particular interest in \(\alpha\) underlines the significance of knowing the expansion property for proper mechanical fitting upon temperature variation. Thermal expansion can lead to practical concerns like the fit of machinery parts, bridge expansion, and even the tightness of jar lids. Students studying this concept gain insight into the practical implications of material selection and design considerations for thermal environments.